Problem: Divide the following complex numbers. $ \dfrac{18-4i}{-3-5i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-3+5i}$ $ \dfrac{18-4i}{-3-5i} = \dfrac{18-4i}{-3-5i} \cdot \dfrac{{-3+5i}}{{-3+5i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(18-4i) \cdot (-3+5i)} {(-3-5i) \cdot (-3+5i)} = \dfrac{(18-4i) \cdot (-3+5i)} {(-3)^2 - (-5i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(18-4i) \cdot (-3+5i)} {(-3)^2 - (-5i)^2} = $ $ \dfrac{(18-4i) \cdot (-3+5i)} {9 + 25} = $ $ \dfrac{(18-4i) \cdot (-3+5i)} {34} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({18-4i}) \cdot ({-3+5i})} {34} = $ $ \dfrac{{18} \cdot {(-3)} + {-4} \cdot {(-3) i} + {18} \cdot {5 i} + {-4} \cdot {5 i^2}} {34} $ Evaluate each product of two numbers. $ \dfrac{-54 + 12i + 90i - 20 i^2} {34} $ Finally, simplify the fraction. $ \dfrac{-54 + 12i + 90i + 20} {34} = \dfrac{-34 + 102i} {34} = -1+3i $